The Fair-Luke algorithm, as appears in Rational Approximations to the Solution of the Second Order Riccati Equation, constructs rational functions which approximate solutions of the so-called “generalized second order Riccati equation” $ $ (A_0+B_0y)y”+(C_0+D_0y)y’-2B_0(y’)^2+E_0+F_0 y+G_0 y^2+H_0y^3=0 \tag{1}$ $ where the coefficients have the form $ $ \begin{align} &A_0=x^2 \sum a_k x^k, \quad B_0=x^2 \sum b_k x^k, \quad C_0=x \sum c_k x^k, \ &D_0=x \sum d_k x^k, \quad E_0=\sum e_k x^k,\; e_0 \neq 0, \quad F_0=\sum f_k x^k,\; f_0 \neq 0 \ &G_0=x \sum g_k x^k, \text{ and } H_0=x \sum h_k x^k, \end{align}$ $ where in all sums $ k$ goes from $ 0$ to $ \infty$ . The key idea is that if we let $ y=y_0$ , and iterate the change of variable $ $ y_n=\frac{\alpha_n}{1+x y_{n+1}} $ $ we obtain a generalized second order Riccati equation for $ y_{n+1}$ , of the form $ $ (A_{n+1}+B_{n+1}y_{n+1})y_{n+1}”+(C_{n+1}+D_{n+1}y_{n+1})y_{n+1}’-2B_{n+1}(y_{n+1}’)^2+E_{n+1}+F_{n+1} y_{n+1}+G_{n+1} y_{n+1}^2+H_{n+1}y_{n+1}^3=0 \tag{2}$ $ where the new coefficients satisfy the algebraic formulas $ $ \begin{align} &A_{n+1}=-A_n-\alpha_n B_n, \ & B_{n+1}=-x A_n, \ & C_{n+1}=-2x^{-1} \left( A_n+\alpha_n B_n \right)-C_n-\alpha_n D_n, \ & D_{n+1}=2A_n-xC_n, \ & E_{n+1}=x^{-1} \left( \alpha_n^{-1} E_n+F_n+\alpha_n G_n+\alpha_n^2 H_n \right), \ &F_{n+1}=-x^{-1} \left( C_n + \alpha_n D_n \right)+3 \alpha_n^{-1} E_n+2F_n+\alpha_n G_n, \ &G_{n+1}=2x^{-1} A_n-C_n+3\alpha_n^{-1} x E_n+x F_n, \ &H_{n+1}=\alpha_n^{-1} x^2 E_n \end{align} $ $

The authors say that the sequence $ (\alpha_n)$ is computed as $ $ y_{n+1}(0)=\alpha_{n+1}=-\frac{E_{n+1}(0)}{F_{n+1}(0)}, \; n=0,1,2,\dots, \tag{3} $ $ and the only way this makes sense to me is if $ $ \alpha_0=-\frac{E_0(0)}{F_0(0)}$ $ as well, for otherwise, the quotient in Equation $ (3)$ would be undefined for $ n=0$ .

I’ve tried applying the algorithm to the first Painlevé equation $ $ u”=6u^2-x, $ $ making use of the change of variable $ u=1+x^2 y$ to get the generalized Riccati equation $ $ x^2 y”(x)+4 x y'(x)-6 x^4 y(x)^2+\left(2-12 x^2\right) y(x)+x-6=0,$ $ with $ \alpha_0=3$ . After one iteration I got the coefficients $ $ A_1=-x^2,B_1=-x^3,C_1=-6 x,D_1=-2 x^2,E_1=-18 x^3-12 x,F_1=-18 x^4+2 \left(2-12 x^2\right)-10,G_1=x \left(-12 x^2-4\right)-2 x,H_1=-2 x^2,\alpha_1=0 $ $ which causes a problem in the second iteration (in particular, $ E_2,F_2,G_2,H_2$ all turn out to be infinite).

My questions are:

- Have I implemented the algorithm correctly? If not, where have I gone wrong?
- It appears that the method isn’t uniformly valid for all initial value problems for the first Painlevé equation. In particular, setting $ u=u_0+u_1x+x^2 y$ results in a generalized second order Riccati equation if and only if $ u_0 \neq 0$ . Is this right?
- Aside from these two points, I’d appreciate any reference for algorithms for solving nonlinear IVPs, where the independent variable is complex, when poles are present (such as in the case of the Painlevé equations).